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I hate this code, but it is faster.

You said you are only working with strings made of characters from a-z. That means 26 characters. The idea is to use 26 counters, each belonging to a character from a-z.

Go through the first(second) string, and every time you find a character c, you increment(decrement) c's counter by 1.

If every counter is zero, that means the strings are permutations of each other, i.e. it's a draw.

Otherwise if every counter is non-negative, you'll know that the first string won; if every counter is non-positive, you'll know that the second string won.

Finally, if some counters have different signs, then it's again a draw.

private static int decide(String sa, String sb){
    int [] charset = new int[26];

    for(int i = 0; i < sa.length(); charset[sa.charAt(i) - 97] ++, i++);
    for(int i = 0; i < sb.length(); charset[sb.charAt(i) - 97] --, i++);

    boolean allnonneg = true;
    boolean allnonpos = true;

    for(int i = 0; i < 25; i++) {
        allnonneg &= (charset[i] >= 0);
        allnonpos &= (charset[i] <= 0);
    }

    if(allnonneg && allnonpos)
        return 0; // every counter 0 => draw
    if(allnonneg)
        return 1; // 1st won
    if(allnonpos)
        return 2; // 2nd won
    else
        return 0; // mixed signs => draw
}